WMC 2018 Senior Problem 34

Level 2

Five circles of equal radius are placed inside a unit square (each side $=1$ ), in such a way so that no two circles intersect in more than $1$ point. Find the largest possible value of the radius.

\frac{\sqrt5-2}{2}

\frac{1+\sqrt2}{5}

\frac{2-\sqrt2}{2}

\frac{\sqrt2-1}{2}

\frac{5-\sqrt5}{5}

1 solution

Julian Yu
Dec 3, 2018

From the image, the diagonal of the square is $4r+2r\sqrt{2}=\sqrt{2}$ . Therefore, $(4+2\sqrt{2})r=\sqrt{2}\implies r=\dfrac{\sqrt{2}}{4+2\sqrt{2}}=\dfrac{\sqrt{2}-1}{2}$ .

WMC 2018 Senior Problem 34

1 solution

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